package 牛客.树;

/*
 * @Author   helen
 * @Date     2021/4/12 16:20
 * @Descriptions
 * 输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。
 * 假设输入的前序遍历和中序遍历的结果中都不含重复的数字。
 * 例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。
        示例1
            输入     [1,2,3,4,5,6,7],[3,2,4,1,6,5,7]
            返回值   {1,2,5,3,4,6,7}
 */

import java.util.Arrays;

public class 重建二叉树 {
    //法一：
    public BTreeNode<Integer> reBuild(int[] pre, int[] in){
        if(pre == null || pre.length == 0)
            return null;
        BTreeNode<Integer> root = new BTreeNode<>(pre[0]);

        int flag = 0;

        for (int i = 0; i < in.length; i++) {
            if(pre[0] == in[i]){
                flag = i;
                break;
            }
        }

        root.left = reBuild(Arrays.copyOfRange(pre, 1,flag+1), Arrays.copyOfRange(in, 0, flag));
        root.right = reBuild(Arrays.copyOfRange(pre, flag+1,pre.length), Arrays.copyOfRange(in, flag+1, in.length));

        return root;

    }

    //法二：
    public BTreeNode<Integer> reConstructBinaryTree(int [] pre, int [] in) {
        if(pre == null || in == null)
            return null;
        BTreeNode<Integer> root = reBuildBTreeNode(pre,0,pre.length - 1, in, 0, in.length - 1);
        return root;
    }

    public BTreeNode<Integer> reBuildBTreeNode(int [] pre, int preStart, int preEnd, int[] in, int inStart, int inEnd){
        if(preStart > preEnd || inStart > inEnd)
            return null;
        BTreeNode<Integer> root = new BTreeNode<>(pre[preStart]);
        for (int i = inStart; i < inEnd; i++) {
            if(in[i] == root.data){
                   //i - inStart为左子树结点个数
                //左子树
                root.left = reBuildBTreeNode(pre,preStart +1, preStart + (i - inStart), in, inStart, i - 1);
                //右子树
                root.right = reBuildBTreeNode(pre, preStart + i - inStart + 1, preEnd, in, i +1, inEnd);
            }
        }
        return root;
    }

    public static void main(String[] args) {
        重建二叉树 obj = new 重建二叉树();
        int[] pre = {1,2,3,4,5,6,7}, in = {3,2,4,1,6,5,7};
        BTreeNode<Integer> root = obj.reConstructBinaryTree(pre, in);
        //BTreeNode<Integer> root = obj.reBuild(pre, in);
        System.out.println(root);
    }
}
